a) \(a^4-5a^2+4=\)\(\left(a^4-4a^2\right)-\left(a^2-4\right)=a^2\left(a^2-4\right)-\left(a^2-4\right)=\left(a^2-1\right)\left(a^2-4\right)\)

\(=\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)\)

\(a^4-a^2+4a-4=a^2\left(a^2-1\right)+4\left(a-1\right)=a^2\left(a-1\right)\left(a+1\right)+4\left(a-1\right)\)

\(=\left(a-1\right)\left[a^2\left(a+1\right)+4\right]=\left(a-1\right)\left(a^3+a^2+4\right)\)

\(a^3+a^2+4=\left(a^3+2a^2\right)-\left(a^2+2a\right)+\left(2a+4\right)=a^2\left(a+2\right)-a\left(a+2\right)+2\left(a+2\right)\)

\(=\left(a^2-a+2\right)\left(a+2\right)\)

\(N=\frac{\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)}{\left(a-1\right)\left(a+2\right)\left(a^2-a+2\right)}=\frac{\left(a+1\right)\left(a-2\right)}{a^2-a+2}\)

em chịu

\(A=\frac{a+b}{a^3+b^3}=\frac{a+b}{\left(a+b\right)\left(a^2-ab+b^2\right)}=\frac{1}{a^2-ab+b^2}\)

\(C=\frac{2ab-b}{8a^3-1}=\frac{b\left(2a-1\right)}{\left(2a-1\right)\left(4a^2+2a+1\right)}=\frac{b}{4a^2+2a+1}\)

Câu b xem lại đề đi nhé

cảm ơn bạn nhé

c)\(P=\)\(\frac{\left(a-b\right)^2-c^2}{\left(a-b+c\right)^2}=\frac{\left(a-b+c\right)\left(a-b-c\right)}{\left(a-b+c\right)^2}=\frac{a-b-c}{a-b+c}\)

b)\(M\)\(=\frac{\left(a+2\right)\left(a-1\right)^2}{\left(2a-3\right)\left(a-1\right)^2}=\frac{a+2}{2a-3}\)

Hắc hắc :P Cứ làm từ từ sẽ thành công em ạ :D

\(=\frac{a+b+a-b}{a^2-b^2}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)

\(=\frac{2a\left(a^2+b^2\right)+2a\left(a^2-b^2\right)}{a^4-b^4}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)

\(=\frac{4a^3\left(a^4+b^4\right)+4a^3\left(a^4-b^4\right)}{a^8-b^8}+\frac{8a^7}{a^8+b^8}\)

\(=\frac{8a^7\left(a^8+b^8\right)+8a^7\left(a^8-b^8\right)}{\left(a^8-b^8\right)\left(a^8+b^8\right)}\)

\(=\frac{16a^{15}}{a^{16}-b^{16}}\)

ĐKXĐ: \(a\ne\frac{3}{2},a\ne-\frac{3}{2}\)

a, \(P=\left(\frac{a-1}{2a-3}-\frac{3a}{4a+6}+\frac{7a-2a^2-1}{18-8a^2}\right):\frac{1}{6-4a}\)

\(=\left(\frac{a-1}{2a-3}-\frac{3a}{2\left(2x+3\right)}+\frac{7a-2a^2-1}{2\left(9-4a^2\right)}\right):\frac{-1}{4a-6}\)

\(=\left(\frac{a-1}{2a-3}-\frac{3a}{2\left(2x+3\right)}-\frac{7a-2a^2-1}{2\left(4a^2-9\right)}\right):\frac{-1}{2\left(2a-3\right)}\)

\(=\left(\frac{a-1}{2a-3}-\frac{3a}{2\left(2x+3\right)}-\frac{7a-2a^2-1}{2\left(2a-3\right)\left(2a+3\right)}\right)\left[-2\left(2a-3\right)\right]\)

\(=\left[\frac{2\left(a-1\right)\left(2a+3\right)-3a\left(2a-3\right)-\left(7a-2a^2-1\right)}{2\left(2a-3\right)\left(2a+3\right)}\right]\left[-2\left(2a-3\right)\right]\)

\(=\frac{4a-5}{2\left(2a-3\right)\left(2a+3\right)}\left[-2\left(2a-3\right)\right]\)

\(=-\frac{\left(4a-5\right)}{2a+3}=\frac{5-4a}{2a+3}\)

a) \(ĐKXĐ:\hept{\begin{cases}a\ne\pm2\\a\ne1\\a\ne0\end{cases}}\)

\(A=\left(\frac{4a}{2+a}+\frac{8a^2}{4-a^2}\right):\left(\frac{a-3}{a^2-2a}-\frac{2}{a}\right)\)

\(\Leftrightarrow A=\frac{8a-4a^2+8a^2}{\left(2-a\right)\left(2+a\right)}:\frac{a-3-2a+4}{a\left(a-2\right)}\)

\(\Leftrightarrow A=\frac{4a^2+8a}{\left(2-a\right)\left(2+a\right)}:\frac{-a+1}{a\left(a-2\right)}\)

\(\Leftrightarrow A=\frac{4a}{2-a}:\frac{-a+1}{a\left(a-2\right)}\)

\(\Leftrightarrow A=\frac{4a^2\left(a-2\right)}{\left(a-2\right)\left(a-1\right)}\)

\(\Leftrightarrow A=\frac{4a^2}{a-1}\)

b) Để A nhận giá trị nguyên

\(\Leftrightarrow\frac{4a^2}{a-1}\inℤ\)

\(\Leftrightarrow4a^2⋮a-1\)

\(\Leftrightarrow4\left(a^2-1\right)+4⋮a-1\)

\(\Leftrightarrow4\left(a-1\right)\left(a+1\right)+4⋮a-1\)

\(\Leftrightarrow4⋮a-1\)

\(\Leftrightarrow a-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

\(\Leftrightarrow a\in\left\{0;2;-1;3;-3;5\right\}\)

Ta sẽ loại các giá trị ở đkxđ

Vậy để \(A\inℤ\Leftrightarrow a\in\left\{2;-1;3;-3;5\right\}\)